Answer:6Step-by-step explanation:The correct notation is: Summation from n = 1 to infinity of the expression:[tex]4(\frac{1}{3} )^{n-1}[/tex]If we expand this notation, we will find that the given summation represent the terms of a G.P and we have to find the sum of infinite terms of a G.P. By placing n=1,2,3,4,... we can find the first few terms as shown below:[tex]4(\frac{1}{3} )^{1-1}=4\\\\ 4(\frac{1}{3} )^{2-1}=4(\frac{1}{3} ) \\\\ 4(\frac{1}{3} )^{3-1} =4(\frac{1}{3} )^{2}[/tex]Its obvious from the above result, that the given summation represent a G.P with first term equal to 4 and common ratio of 1/3. The formula of infinite G.P is:[tex]S=\frac{a_{1} }{1-r}[/tex]Using the value of a1 = 4 and r = 1/3 we can calculate the given summation[tex]S=\frac{4}{1-\frac{1}{3} }=6[/tex]Therefore, the given summation is equal to 6.